Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
I got the same error (TypeError: 'int' object is not callable)
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x(1-x))
return xl
... ... ... ...
>>> xlim(1,100,0,0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in xlim
TypeError: 'int' object is not callable
after reading this post I realized that I forgot a multiplication sign * so
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x * (1-x))
return xl
xlim(1.0,100.0,0.0,0.0)
0.005
tanks
Stop stomping on round somewhere else by binding an int to it.
I was also facing this issue but in a little different scenario.
Scenario:
param = 1
def param():
.....
def func():
if param:
var = {passing a dict here}
param(var)
It looks simple and a stupid mistake here, but due to multiple lines of codes in the actual code, it took some time for me to figure out that the variable name I was using was same as my function name because of which I was getting this error.
Changed function name to something else and it worked.
So, basically, according to what I understood, this error means that you are trying to use an integer as a function or in more simple terms, the called function name is also used as an integer somewhere in the code.
So, just try to find out all occurrences of the called function name and look if that is being used as an integer somewhere.
I struggled to find this, so, sharing it here so that someone else may save their time, in case if they get into this issue.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
Sometimes the problem would be forgetting an operator while calculation.
Example:
print(n-(-1+(math.sqrt(1-4(2*(-n))))/2)) rather
it has to be
print(n-(-1+(math.sqrt(1-4*(2*(-n))))/2))
HTH
There are two reasons for this error "TypeError: 'int' object is not callable"
Function Has an Integer Value
Consider
a = [5, 10, 15, 20]
max = 0
max = max(a)
print(max)
This will produce TypeError: 'int' object is not callable.
Just change the variable name "max" to var(say).
a = [5, 10, 15, 20]
var = 0
var = max(a)
print(var)
The above code will run perfectly without any error!!
Missing a Mathematical Operator
Consider
a = 5
b = a(a+1)
print(b)
This will also produce TypeError: 'int' object is not callable.
You might have forgotten to put the operator in between ( '*' in this case )
As mentioned you might have a variable named round (of type int) in your code and removing that should get rid of the error. For Jupyter notebooks however, simply clearing a cell or deleting it might not take the variable out of scope. In such a case, you can restart your notebook to start afresh after deleting the variable.
You can always use the below method to disambiguate the function.
__import__('__builtin__').round((a/b)*0.9*c)
__builtin__ is the module name for all the built in functions like round, min, max etc. Use the appropriate module name for functions from other modules.
I encountered this error because I was calling a function inside my model that used the #property decorator.
#property
def volume_range(self):
return self.max_oz - self.min_oz
When I tried to call this method in my serializer, I hit the error "TypeError: 'int' object is not callable".
def get_oz_range(self, obj):
return obj.volume_range()
In short, the issue was that the #property decorator turns a function into a getter. You can read more about property() in this SO response.
The solution for me was to access volume_range like a variable and not call it as a function:
def get_oz_range(self, obj):
return obj.volume_range # No more parenthesis
Related
Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
I got the same error (TypeError: 'int' object is not callable)
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x(1-x))
return xl
... ... ... ...
>>> xlim(1,100,0,0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in xlim
TypeError: 'int' object is not callable
after reading this post I realized that I forgot a multiplication sign * so
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x * (1-x))
return xl
xlim(1.0,100.0,0.0,0.0)
0.005
tanks
Stop stomping on round somewhere else by binding an int to it.
I was also facing this issue but in a little different scenario.
Scenario:
param = 1
def param():
.....
def func():
if param:
var = {passing a dict here}
param(var)
It looks simple and a stupid mistake here, but due to multiple lines of codes in the actual code, it took some time for me to figure out that the variable name I was using was same as my function name because of which I was getting this error.
Changed function name to something else and it worked.
So, basically, according to what I understood, this error means that you are trying to use an integer as a function or in more simple terms, the called function name is also used as an integer somewhere in the code.
So, just try to find out all occurrences of the called function name and look if that is being used as an integer somewhere.
I struggled to find this, so, sharing it here so that someone else may save their time, in case if they get into this issue.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
Sometimes the problem would be forgetting an operator while calculation.
Example:
print(n-(-1+(math.sqrt(1-4(2*(-n))))/2)) rather
it has to be
print(n-(-1+(math.sqrt(1-4*(2*(-n))))/2))
HTH
There are two reasons for this error "TypeError: 'int' object is not callable"
Function Has an Integer Value
Consider
a = [5, 10, 15, 20]
max = 0
max = max(a)
print(max)
This will produce TypeError: 'int' object is not callable.
Just change the variable name "max" to var(say).
a = [5, 10, 15, 20]
var = 0
var = max(a)
print(var)
The above code will run perfectly without any error!!
Missing a Mathematical Operator
Consider
a = 5
b = a(a+1)
print(b)
This will also produce TypeError: 'int' object is not callable.
You might have forgotten to put the operator in between ( '*' in this case )
As mentioned you might have a variable named round (of type int) in your code and removing that should get rid of the error. For Jupyter notebooks however, simply clearing a cell or deleting it might not take the variable out of scope. In such a case, you can restart your notebook to start afresh after deleting the variable.
You can always use the below method to disambiguate the function.
__import__('__builtin__').round((a/b)*0.9*c)
__builtin__ is the module name for all the built in functions like round, min, max etc. Use the appropriate module name for functions from other modules.
I encountered this error because I was calling a function inside my model that used the #property decorator.
#property
def volume_range(self):
return self.max_oz - self.min_oz
When I tried to call this method in my serializer, I hit the error "TypeError: 'int' object is not callable".
def get_oz_range(self, obj):
return obj.volume_range()
In short, the issue was that the #property decorator turns a function into a getter. You can read more about property() in this SO response.
The solution for me was to access volume_range like a variable and not call it as a function:
def get_oz_range(self, obj):
return obj.volume_range # No more parenthesis
Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
I got the same error (TypeError: 'int' object is not callable)
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x(1-x))
return xl
... ... ... ...
>>> xlim(1,100,0,0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in xlim
TypeError: 'int' object is not callable
after reading this post I realized that I forgot a multiplication sign * so
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x * (1-x))
return xl
xlim(1.0,100.0,0.0,0.0)
0.005
tanks
Stop stomping on round somewhere else by binding an int to it.
I was also facing this issue but in a little different scenario.
Scenario:
param = 1
def param():
.....
def func():
if param:
var = {passing a dict here}
param(var)
It looks simple and a stupid mistake here, but due to multiple lines of codes in the actual code, it took some time for me to figure out that the variable name I was using was same as my function name because of which I was getting this error.
Changed function name to something else and it worked.
So, basically, according to what I understood, this error means that you are trying to use an integer as a function or in more simple terms, the called function name is also used as an integer somewhere in the code.
So, just try to find out all occurrences of the called function name and look if that is being used as an integer somewhere.
I struggled to find this, so, sharing it here so that someone else may save their time, in case if they get into this issue.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
Sometimes the problem would be forgetting an operator while calculation.
Example:
print(n-(-1+(math.sqrt(1-4(2*(-n))))/2)) rather
it has to be
print(n-(-1+(math.sqrt(1-4*(2*(-n))))/2))
HTH
There are two reasons for this error "TypeError: 'int' object is not callable"
Function Has an Integer Value
Consider
a = [5, 10, 15, 20]
max = 0
max = max(a)
print(max)
This will produce TypeError: 'int' object is not callable.
Just change the variable name "max" to var(say).
a = [5, 10, 15, 20]
var = 0
var = max(a)
print(var)
The above code will run perfectly without any error!!
Missing a Mathematical Operator
Consider
a = 5
b = a(a+1)
print(b)
This will also produce TypeError: 'int' object is not callable.
You might have forgotten to put the operator in between ( '*' in this case )
As mentioned you might have a variable named round (of type int) in your code and removing that should get rid of the error. For Jupyter notebooks however, simply clearing a cell or deleting it might not take the variable out of scope. In such a case, you can restart your notebook to start afresh after deleting the variable.
You can always use the below method to disambiguate the function.
__import__('__builtin__').round((a/b)*0.9*c)
__builtin__ is the module name for all the built in functions like round, min, max etc. Use the appropriate module name for functions from other modules.
I encountered this error because I was calling a function inside my model that used the #property decorator.
#property
def volume_range(self):
return self.max_oz - self.min_oz
When I tried to call this method in my serializer, I hit the error "TypeError: 'int' object is not callable".
def get_oz_range(self, obj):
return obj.volume_range()
In short, the issue was that the #property decorator turns a function into a getter. You can read more about property() in this SO response.
The solution for me was to access volume_range like a variable and not call it as a function:
def get_oz_range(self, obj):
return obj.volume_range # No more parenthesis
This question already has answers here:
Why does code like `str = str(...)` cause a TypeError, but only the second time?
(20 answers)
Closed last month.
I tried to use this code from a tutorial at the REPL:
example = list('easyhoss')
The tutorial says that example should become equal to a list ['e', 'a', 's', 'y', 'h', 'o', 's', 's']. But I got an error instead:
>>> example = list('easyhoss')
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'list' object is not callable
Why did this happen?
Seems like you've shadowed the builtin name list, which points at a class, by the same name pointing at an instance of it. Here is an example:
>>> example = list('easyhoss') # here `list` refers to the builtin class
>>> list = list('abc') # we create a variable `list` referencing an instance of `list`
>>> example = list('easyhoss') # here `list` refers to the instance
Traceback (most recent call last):
File "<string>", line 1, in <module>
TypeError: 'list' object is not callable
I believe this is fairly obvious. Python stores object names (functions and classes are objects, too) in namespaces (which are implemented as dictionaries), hence you can rewrite pretty much any name in any scope. It won't show up as an error of some sort. As you might know, Python emphasizes that "special cases aren't special enough to break the rules". And there are two major rules behind the problem you've faced:
Namespaces. Python supports nested namespaces. Theoretically you can endlessly nest them. As I've already mentioned, they are basically dictionaries of names and references to corresponding objects. Any module you create gets its own "global" namespace, though in fact it's just a local namespace with respect to that particular module.
Scoping. When you reference a name, the Python runtime looks it up in the local namespace (with respect to the reference) and, if such name does not exist, it repeats the attempt in a higher-level namespace. This process continues until there are no higher namespaces left. In that case you get a NameError. Builtin functions and classes reside in a special high-order namespace __builtins__. If you declare a variable named list in your module's global namespace, the interpreter will never search for that name in a higher-level namespace (that is __builtins__). Similarly, suppose you create a variable var inside a function in your module, and another variable var in the module. Then, if you reference var inside the function, you will never get the global var, because there is a var in the local namespace - the interpreter has no need to search it elsewhere.
Here is a simple illustration.
>>> example = list("abc") # Works fine
>>>
>>> # Creating name "list" in the global namespace of the module
>>> list = list("abc")
>>>
>>> example = list("abc")
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: 'list' object is not callable
>>> # Python looks for "list" and finds it in the global namespace,
>>> # but it's not the proper "list".
>>>
>>> # Let's remove "list" from the global namespace
>>> del list
>>> # Since there is no "list" in the global namespace of the module,
>>> # Python goes to a higher-level namespace to find the name.
>>> example = list("abc") # It works.
So, as you see there is nothing special about Python builtins. And your case is a mere example of universal rules. You'd better use an IDE (e.g. a free version of PyCharm, or Atom with Python plugins) that highlights name shadowing to avoid such errors.
You might as well be wondering what is a "callable", in which case you can read this post. list, being a class, is callable. Calling a class triggers instance construction and initialisation. An instance might as well be callable, but list instances are not. If you are even more puzzled by the distinction between classes and instances, then you might want to read the documentation (quite conveniently, the same page covers namespaces and scoping).
If you want to know more about builtins, please read the answer by Christian Dean.
P.S. When you start an interactive Python session, you create a temporary module.
Before you can fully understand what the error means and how to solve, it is important to understand what a built-in name is in Python.
What is a built-in name?
In Python, a built-in name is a name that the Python interpreter already has assigned a predefined value. The value can be either a function or class object. These names are always made available by default, no matter the scope. Some of the values assigned to these names represent fundamental types of the Python language, while others are simple useful.
As of the latest version of Python - 3.6.2 - there are currently 61 built-in names. A full list of the names and how they should be used, can be found in the documentation section Built-in Functions.
An important point to note however, is that Python will not stop you from re-assigning builtin names. Built-in names are not reserved, and Python allows them to be used as variable names as well.
Here is an example using the dict built-in:
>>> dict = {}
>>> dict
{}
>>>
As you can see, Python allowed us to assign the dict name, to reference a dictionary object.
What does "TypeError: 'list' object is not callable" mean?
To put it simply, the reason the error is occurring is because you re-assigned the builtin name list in the script:
list = [1, 2, 3, 4, 5]
When you did this, you overwrote the predefined value of the built-in name. This means you can no longer use the predefined value of list, which is a class object representing Python list.
Thus, when you tried to use the list class to create a new list from a range object:
myrange = list(range(1, 10))
Python raised an error. The reason the error says "'list' object is not callable", is because as said above, the name list was referring to a list object. So the above would be the equivalent of doing:
[1, 2, 3, 4, 5](range(1, 10))
Which of course makes no sense. You cannot call a list object.
How can I fix the error?
Suppose you have code such as the following:
list = [1, 2, 3, 4, 5]
myrange = list(range(1, 10))
for number in list:
if number in myrange:
print(number, 'is between 1 and 10')
Running the above code produces the following error:
Traceback (most recent call last):
File "python", line 2, in <module>
TypeError: 'list' object is not callable
If you are getting a similar error such as the one above saying an "object is not callable", chances are you used a builtin name as a variable in your code. In this case and other cases the fix is as simple as renaming the offending variable. For example, to fix the above code, we could rename our list variable to ints:
ints = [1, 2, 3, 4, 5] # Rename "list" to "ints"
myrange = list(range(1, 10))
for number in ints: # Renamed "list" to "ints"
if number in myrange:
print(number, 'is between 1 and 10')
PEP8 - the official Python style guide - includes many recommendations on naming variables.
This is a very common error new and old Python users make. This is why it's important to always avoid using built-in names as variables such as str, dict, list, range, etc.
Many linters and IDEs will warn you when you attempt to use a built-in name as a variable. If your frequently make this mistake, it may be worth your time to invest in one of these programs.
I didn't rename a built-in name, but I'm still getting "TypeError: 'list' object is not callable". What gives?
Another common cause for the above error is attempting to index a list using parenthesis (()) rather than square brackets ([]). For example:
>>> lst = [1, 2]
>>> lst(0)
Traceback (most recent call last):
File "<pyshell#32>", line 1, in <module>
lst(0)
TypeError: 'list' object is not callable
For an explanation of the full problem and what can be done to fix it, see TypeError: 'list' object is not callable while trying to access a list.
If you are in a interactive session and don't want to restart you can remove the shadowing with
del list
In the league of stupid Monday morning mistakes, using round brackets instead of square brackets when trying to access an item in the list will also give you the same error message:
l=[1,2,3]
print(l[2])#GOOD
print(l(2))#BAD
TypeError: 'list' object is not callable
You may have used built-in name 'list' for a variable in your code.
If you are using Jupyter notebook, sometimes even if you change the name of that variable from 'list' to something different and rerun that cell, you may still get the error. In this case you need to restart the Kernal.
In order to make sure that the name has change, click on the word 'list' when you are creating a list object and press Shift+Tab, and check if Docstring shows it as an empty list.
Why does TypeError: 'list' object is not callable appear?
Explanation:
It is because you defined list as a variable before (i am pretty sure), so it would be a list, not the function anymore, that's why everyone shouldn't name variables functions, the below is the same as what you're doing now:
>>> [1,2,3]()
Traceback (most recent call last):
File "<pyshell#0>", line 1, in <module>
[1,2,3]()
TypeError: 'list' object is not callable
>>>
So you need it to be the default function of list, how to detect if it is? just use:
>>> list
<class 'list'>
>>> list = [1,2,3]
>>> list
[1, 2, 3]
>>> list()
Traceback (most recent call last):
File "<pyshell#4>", line 1, in <module>
list()
TypeError: 'list' object is not callable
>>>
How do i detect whether a variable name is a function? well, just simple see if it has a different color, or use a code like:
>>> 'list' in dir(__builtins__)
True
>>> 'blah' in dir(__builtins__)
False
>>>
After this, you should know why does TypeError: 'list' object is not callable appear.
Okay, so now...
How to fix this TypeError: 'list' object is not callable error?
Code:
You have to either do __builtins__.list():
>>> list = [1,2,3]
>>> __builtins__.list()
[]
>>>
Or use []:
>>> list = [1,2,3]
>>> []
[]
>>>
Or remove list variable from memory:
>>> list = [1,2,3]
>>> del list
>>> list()
[]
>>>
Or just rename the variable:
>>> lst = [1,2,3]
>>> list()
[]
>>>
P.S. Last one is the most preferable i guess :-)
There are a whole bunch of solutions that work.
References:
'id' is a bad variable name in Python
How do I use a keyword as a variable name?
How to use reserved keyword as the name of variable in python?
find out what you have assigned to 'list' by displaying it
>>> print(list)
if it has content, you have to clean it with
>>> del list
now display 'list' again and expect this
<class 'list'>
Once you see this, you can proceed with your copy.
For me it was a flask server returning some videos array (which I expected to be in json format..)
adding json.dumps(videos) fixed this issue
I was getting this error for another reason:
I accidentally had a blank list created in my __init__ which had the same name as a method I was trying to call (I had just finished refactoring a bit and the variable was no longer needed, but I missed it when cleaning up). So when I was instantiating the class and trying to call the method, it thought I was referencing the list object, not the method:
class DumbMistake:
def __init__(self, k, v):
self.k = k
self.v = v
self.update = []
def update(self):
// do updates to k, v, etc
if __name__ == '__main__':
DumbMistake().update('one,two,three', '1,2,3')
So it was trying to assign the two strings to self.update[] instead of calling the update() method. Removed the variable and it all worked as intended. Hope this helps someone.
You have already assigned a value to list.
So, you cannot use the list() when it’s a variable.
Restart the shell or IDE, by pressing Ctrl+F6 on your computer.
Hope this works too.
I found myself getting this error because I had a method that returned a list that I gave a #property decorator. I forgot about the decorator and was calling method() instead of just method which gave this same error.
Why error occurred?
Because you have named any of your list as "list" in current kernel.
example:
import operator
list = [1, 4, 5, 7, 9, 11]
#you are naming list as "list"
print("The sum is : ", end="")
print(list(itertools.accumulate(list1)))
print("The product is : ", end="")
print(list(itertools.accumulate(list1, operator.mul)))
Solution:
Simply restart the kernel.
Close the current interpreter using exit() command and reopen typing python to start your work. And do not name a list as list literally. Then you will be fine.
to solve the error like this one: "list object is not callable in python" even you are changing the variable name then please restart the kernel in Python Jutyter Notebook if you are using it or simply restart the IDE.
I hope this will work. Thank you!!!
I have some code like:
def example(parameter):
global str
str = str(parameter)
print(str)
example(1)
example(2)
The first call to example works, but then the second time around I get an error like:
Traceback (most recent call last):
File "test.py", line 7, in <module>
example(2)
File "test.py", line 3, in example
str = str(parameter)
TypeError: 'str' object is not callable
Why does this happen, and how can I fix it?
If you are in an interactive session and encountered a problem like this, and you want to fix the problem without restarting the interpreter, see How to restore a builtin that I overwrote by accident?.
Where the code says:
global str
str = str(parameter)
You are redefining what str() means. str is the built-in Python name of the string type, and you don't want to change it.
Use a different name for the local variable, and remove the global statement.
Note that if you used code like this at the Python REPL, then the assignment to the global str will persist until you do something about it. You can restart the interpreter, or del str. The latter works because str is not actually a defined global variable by default - instead, it's normally found in a fallback (the builtins standard library module, which is specially imported at startup and given the global name __builtins__).
While not in your code, another hard-to-spot error is when the % character is missing in an attempt of string formatting:
"foo %s bar %s coffee"("blah","asdf")
but it should be:
"foo %s bar %s coffee"%("blah","asdf")
The missing % would result in the same TypeError: 'str' object is not callable.
In my case I had a class that had a method and a string property of the same name, I was trying to call the method but was getting the string property.
Note that TypeError: 'str' object is not callable means only that there is an attempt to call (i.e., use function-call syntax) a string (i.e., any name that previously had a string assigned to it). Using any other built-in method as variable name can cause the exact same error message.
You can get this error if you have variable str and trying to call str() function.
Whenever that happens, just issue the following ( it was also posted above)
>>> del str
That should fix it.
Another case of this: Messing with the __repr__ function of an object where a format() call fails non-transparently.
In our case, we used a #property decorator on the __repr__ and passed that object to a format(). The #property decorator causes the __repr__ object to be turned into a string, which then results in the str object is not callable error.
Check your input parameters, and make sure you don't have one named type. If so then you will have a clash and get this error.
str = 'Hello World String'
print(str(10)+' Good day!!')
Even I faced this issue with the above code as we are shadowing str() function.
Solution is:
string1 = 'Hello World String'
print(str(10)+' Good day!!')
I had the same error. In my case wasn't because of a variable named str. But because I named a function with a str parameter and the variable the same.
same_name = same_name(var_name: str)
I run it in a loop. The first time it run ok. The second time I got this error. Renaming the variable to a name different from the function name fixed this. So I think it's because Python once associate a function name in a scope, the second time tries to associate the left part (same_name =) as a call to the function and detects that the str parameter is not present, so it's missing, then it throws that error.
This error can also occur as a result of trying to call a property (as though it were a function):
class Example:
#property
def value():
return 'test'
e = Example()
print(e.value()) # should just be `e.value` to get the string
This problem can be caused by code like:
"Foo" ("Bar" if bar else "Baz")
You can concatenate string literals by putting them next to each other, like "Foo" "Bar". However, because of the open parenthesis, the code was interpreted as an attempt to call the string "Foo" as if it were a function.
it could be also you are trying to index in the wrong way:
a = 'apple'
a(3) ===> 'str' object is not callable
a[3] = l
it is recommended not to use str int list etc.. as variable names, even though python will allow it.
this is because it might create such accidents when trying to access reserved keywords that are named the same
This error could also occur with code like:
class Shape:
def __init__(self, colour):
self.colour = colour
def colour(self):
print("colour:", self.colour)
myShape = Shape("pink")
myShape.colour()
In the __init__ method, we assign an attribute colour, which has the same name as the method colour. When we later attempt to call the method, the instance's attribute is looked up instead. myShape.colour is the string "pink", which is not callable.
To fix this, change either the method name or the variable name.
I also got this error.
For me it was just a typo:
I wrote:
driver.find_element_by_id("swal2-content").text()
while it should have been:
driver.find_element_by_id("swal2-content").text
In my case, I had a Class with a method in it. The method did not have 'self' as the first parameter and the error was being thrown when I made a call to the method. Once I added 'self,' to the method's parameter list, it was fine.
FWIW I just hit this on a slightly different use case. I scoured and scoured my code looking for where I might've used a 'str' variable, but could not find it. I started to suspect that maybe one of the modules I imported was the culprit... but alas, it was a missing '%' character in a formatted print statement.
Here's an example:
x=5
y=6
print("x as a string is: %s. y as a string is: %s" (str(x) , str(y)) )
This will result in the output:
TypeError: 'str' object is not callable
The correction is:
x=5
y=6
print("x as a string is: %s. y as a string is: %s" % (str(x) , str(y)) )
Resulting in our expected output:
x as a string is: 5. y as a string is: 6
It also give same error if math library not imported,
import math
I realize this is not a runtime warning, but PyCharm gave me this similarly-worded IDE warning:
if hasattr(w, 'to_json'):
return w.to_json()
# warning, 'str' object is not callable
This was because the IDE assumed w.to_json was a string. The solution was to add a callable() check:
if hasattr(w, 'to_json') and callable(w.to_json):
return w.to_json()
Then the warning went away. This same check may also prevent the runtime exception in the original question.
Given the following integers and calculation
from __future__ import division
a = 23
b = 45
c = 16
round((a/b)*0.9*c)
This results in:
TypeError: 'int' object is not callable.
How can I round the output to an integer?
Somewhere else in your code you have something that looks like this:
round = 42
Then when you write
round((a/b)*0.9*c)
that is interpreted as meaning a function call on the object bound to round, which is an int. And that fails.
The problem is whatever code binds an int to the name round. Find that and remove it.
I got the same error (TypeError: 'int' object is not callable)
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x(1-x))
return xl
... ... ... ...
>>> xlim(1,100,0,0)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "<stdin>", line 3, in xlim
TypeError: 'int' object is not callable
after reading this post I realized that I forgot a multiplication sign * so
def xlim(i,k,s1,s2):
x=i/(2*k)
xl=x*(1-s2*x-s1*(1-x)) / (1-s2*x**2-2*s1*x * (1-x))
return xl
xlim(1.0,100.0,0.0,0.0)
0.005
tanks
Stop stomping on round somewhere else by binding an int to it.
I was also facing this issue but in a little different scenario.
Scenario:
param = 1
def param():
.....
def func():
if param:
var = {passing a dict here}
param(var)
It looks simple and a stupid mistake here, but due to multiple lines of codes in the actual code, it took some time for me to figure out that the variable name I was using was same as my function name because of which I was getting this error.
Changed function name to something else and it worked.
So, basically, according to what I understood, this error means that you are trying to use an integer as a function or in more simple terms, the called function name is also used as an integer somewhere in the code.
So, just try to find out all occurrences of the called function name and look if that is being used as an integer somewhere.
I struggled to find this, so, sharing it here so that someone else may save their time, in case if they get into this issue.
In my case I changed:
return <variable>
with:
return str(<variable>)
try with the following and it must work:
str(round((a/b)*0.9*c))
Sometimes the problem would be forgetting an operator while calculation.
Example:
print(n-(-1+(math.sqrt(1-4(2*(-n))))/2)) rather
it has to be
print(n-(-1+(math.sqrt(1-4*(2*(-n))))/2))
HTH
There are two reasons for this error "TypeError: 'int' object is not callable"
Function Has an Integer Value
Consider
a = [5, 10, 15, 20]
max = 0
max = max(a)
print(max)
This will produce TypeError: 'int' object is not callable.
Just change the variable name "max" to var(say).
a = [5, 10, 15, 20]
var = 0
var = max(a)
print(var)
The above code will run perfectly without any error!!
Missing a Mathematical Operator
Consider
a = 5
b = a(a+1)
print(b)
This will also produce TypeError: 'int' object is not callable.
You might have forgotten to put the operator in between ( '*' in this case )
As mentioned you might have a variable named round (of type int) in your code and removing that should get rid of the error. For Jupyter notebooks however, simply clearing a cell or deleting it might not take the variable out of scope. In such a case, you can restart your notebook to start afresh after deleting the variable.
You can always use the below method to disambiguate the function.
__import__('__builtin__').round((a/b)*0.9*c)
__builtin__ is the module name for all the built in functions like round, min, max etc. Use the appropriate module name for functions from other modules.
I encountered this error because I was calling a function inside my model that used the #property decorator.
#property
def volume_range(self):
return self.max_oz - self.min_oz
When I tried to call this method in my serializer, I hit the error "TypeError: 'int' object is not callable".
def get_oz_range(self, obj):
return obj.volume_range()
In short, the issue was that the #property decorator turns a function into a getter. You can read more about property() in this SO response.
The solution for me was to access volume_range like a variable and not call it as a function:
def get_oz_range(self, obj):
return obj.volume_range # No more parenthesis